# Show that $F(x)$ (an inverse function of $G(x)$) is differentiable and that $F'(x)$ is $F(x)$, not assuming $F(x) = e^x$

Good morning! I sort of feel I have posted far too many questions these past days, but I just think the explanations given here are much more insightful (and helpful) than those in my textbook..

So anyway, if you are willing to lend a hand (neuron?).

I have a function $F(x)$ which is the inverse of the function: \begin{align}G(x) &= \int_{1}^{x}\frac{1}{t}\text{d}t\\ F(x) &= G^{-1}(x)\end{align}

Where \begin{align}V_F &= (0, \infty)\\ D_F &= R\end{align}

What we are looking to prove is that $F(x)$ is differentiable and that $F'(x) = F(x)$. I have seen a few proofs on this, but I don't quite understand why they are proofs (I am aware there are a few threads here regarding the differentiability of inverse functions, but I couldn't seem to wrap my head around them).

Do you know how to differentiate $G$? If yes you should easily see that $$G^\prime(x) = \frac{1}{x}$$ on the other hand you can derive that for any inverse function $F(x) = G^{-1}(x)$ the relation $$F^\prime(x) = \frac{1}{G^\prime(F(x))}$$ holds. If you now insert $G^\prime$ this results in $$F^\prime(x) = \frac{1}{G^\prime(F(x))} = \frac{1}{\frac{1}{F(x)}}= F(x)$$